Difference between revisions of "2012 USAJMO Problems/Problem 3"
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<cmath>\frac{a^3 + 3b^3}{5a + b} + \frac{b^3 + 3c^3}{5b + c} + \frac{c^3 + 3a^3}{5c + a} \ge \frac{1 + 3}{6} (a^2 + b^2 + c^2) = \frac{2}{3} (a^2 + b^2 + c^2).</cmath> | <cmath>\frac{a^3 + 3b^3}{5a + b} + \frac{b^3 + 3c^3}{5b + c} + \frac{c^3 + 3a^3}{5c + a} \ge \frac{1 + 3}{6} (a^2 + b^2 + c^2) = \frac{2}{3} (a^2 + b^2 + c^2).</cmath> | ||
==Solution 2== | ==Solution 2== | ||
| − | Split up the numerators and multiply both sides of the fraction by either a or 3a | + | Titu's Lemma: The sum of multiple fractions in the form <math>\frac{a_n^2}{b_n}</math> where <math>a_n</math> and <math>b_n</math> are sequences of real numbers is greater than of equal to the square of the sum of all <math>a_i</math> divided by the sum of all <math>b_i</math>, where i is a whole number less than n+1. Titu's Lemma can be proved using the Cauchy-Schwarz Inequality after multiplying out the denominator of the RHS. |
| − | < | + | |
| − | Then use Titu's Lemma: < | + | Consider the LHS of the proposed inequality. Split up the numerators and multiply both sides of the fraction by either a or 3a to make the LHS |
| − | + | <cmath>\sum_{cyc} \frac {a^4} {5a^2+ab}+\sum_{cyc} \frac {9a^4} {15ac+3a^2}</cmath> | |
| − | + | (Cyclic notation is a special notation in which all permutations of (a,b,c) is the summation command.) | |
| + | |||
| + | Then use Titu's Lemma on all terms: <cmath>\sum_{cyc} \frac {a^4} {5a^2+ab} +\sum_{cyc} \frac {9a^4} {15ac+3a^2} \ge \frac {16(a^2+b^2+c^2)^2} {8(a^2+b^2+c^2)+16(ab+bc+ca)} \ge \frac {16(a^2 + b^2 + c^2)^2}{24(a^2 + b^2 + c^2)} = \frac{2}{3} (a^2 + b^2 + c^2) </cmath> owing to the fact that <math>a^2+b^2+c^2 \ge ab+bc+ca</math>, which is actually equivalent to <math>(a-b)^2 + (b-c)^2 + (c-a)^2 \ge 0</math>! | ||
==Solution 3== | ==Solution 3== | ||
We proceed to prove that | We proceed to prove that | ||
| − | <cmath>\frac{a^3 + 3b^3}{5a + b} \ge -\frac{1}{36} a^2 + \frac{25}{36} b^2</cmath> | + | <cmath>\frac{a^3 + 3b^3}{5a + b} \ge -\frac{1}{36} a^2 + \frac{25}{36} b^2 </cmath> |
| + | |||
| + | (then the inequality in question is just the cyclic sum of both sides, since <cmath>\sum_{cyc} (-\frac{1}{36} a^2 + \frac{25}{36} b^2) = \frac{24}{36}\sum_{cyc} a^2 = \frac{2}{3} (a^2+b^2+c^2)</cmath> | ||
| + | ) | ||
Indeed, by AP-GP, we have | Indeed, by AP-GP, we have | ||
| − | <cmath> 41 (a^3 + b^3+b^3) \ge 41 | + | <cmath> 41 (a^3 + b^3+b^3) \ge 41 \cdot 3 ab^2 </cmath> |
and | and | ||
| Line 51: | Line 56: | ||
which is equivalent to: | which is equivalent to: | ||
| − | <cmath> 36(a^3 + | + | <cmath> 36(a^3 + 3b^3) \ge (5a + b)(-a^2 + 25b^2) </cmath> |
Dividing <math>36(5a+b)</math> from both sides, the desired inequality is proved. | Dividing <math>36(5a+b)</math> from both sides, the desired inequality is proved. | ||
| − | + | --[[User:Lightest|Lightest]] 15:31, 7 May 2012 (EDT) | |
| + | |||
| + | ==Solution 4== | ||
| + | By Cauchy-Schwarz, | ||
| − | <cmath>\ | + | <cmath>\left(\sum_{cyc} \dfrac{a^3}{5a + b} + \dfrac{b^3}{5a + b} + \dfrac{b^3}{5a + b} + \dfrac{b^3}{5a + b} \right) \ge \dfrac{\left( \sum_{cyc} a^2 + b^2 + b^2 + b^2 \right)^2}{ \left( \sum_{cyc} a(5a + b) + b(5a + b) + b(5a + b) + b(5a + b) \right) }</cmath> |
| − | <cmath>\ | + | <cmath>= \dfrac{\left(4a^2 + 4b^2 + 4c^2\right)^2}{\left(8a^2 + 8b^2 + 8c^2\right) + \left(16ab + 16bc + 16ca\right)}</cmath> |
| − | <cmath>\ | + | <cmath>\ge \dfrac{16\left(a^2 + b^2 + c^2\right)^2}{\left(8a^2 + 8b^2 + 8c^2\right) + \left(8a^2 + 8b^2\right) + \left(8b^2 + 8c^2\right) + \left(8c^2 + 8a^2\right)}</cmath> (by AM-GM) <cmath>= \dfrac{2}{3}\left(a^2 + b^2 + c^2\right)</cmath> as desired. |
| − | + | ==Solution 5== | |
| − | -- | + | We note that if we can prove that there exists a real number <math>x</math> such that <cmath>\frac{a^3+3b^3}{5a+b}\geq xa^2+(\frac{2}{3}-x)b^2</cmath> for all positive reals <math>a</math> and <math>b</math>, then we are done since both sides are a cyclic sum of <math>a,b</math>, and <math>c</math>. Note that if we multiply both <math>a</math> and <math>b</math> by a constant <math>r</math>, the left and right sides will both multiply by <math>r^2</math> (since the numerator on the left will multiply by <math>r^3</math> and the denominator will multiply by <math>r</math>. This means that if the inequality is satisfied for an ordered pair <math>(a,b)</math>, then it is also satisfied for <math>(ar,br)</math> for any positive real <math>r</math>. Thus, without loss of generality we can let <math>b=1</math>,and our inequality becomes <cmath>\frac{a^3+3}{5a+1}\geq a^2x+\frac{2}{3}-x.</cmath> Expanding this and rearranging we get <math>a^3(1-5x)+a^2(-x)+a(5x-\frac{10}{3})+x+\frac{7}{3}\geq0</math>. Since <math>1</math> is a root of the left side, we can factor this as <cmath>(a-1)(a^2(1-5x)+a(1-6x)-x-\frac{7}{3})\geq0.</cmath> Note that for this to always be nonnegative, we must have <math>a^2(1-5x)+a(1-6x)-x-\frac{7}{3}</math> be positive when <math>a>1</math> and negative when <math>a<1</math>. Thus, it must have a root at <math>a=1</math>, so we plug in <math>a=1</math> is a root and find that we must have <math>x=-\frac{1}{36}</math>. When <math>x=-\frac{1}{36}</math>, the expression factors as <math>(a-1)^2(\frac{41}{36}a+\frac{83}{36})\geq0</math> which obviously is nonnegative for all positive reals <math>a</math>, so we are done. |
| + | |||
| + | ~john0512, bronzetruck2016 | ||
| + | |||
| + | |||
| + | ==Solution 6== | ||
| + | |||
| + | <math>LHS=\sum_{cyc}\frac{a^3+3b^3}{5a+b}=\sum_{cyc}\frac{a^3}{5a+b}+\sum_{cyc}\frac{3b^3}{5a+b}=\sum_{cyc}\frac{a^4}{5a^2+ab}+\sum_{cyc}\frac{9b^4}{15ab+3b^2}</math> which tells that <math>LHS\geq \frac{(a^2+b^2+c^2)^2}{5(a^2+b^2+c^2)+ab+bc+ac}+\frac{3(a^2+b^2+c^2)}{(a^2+b^2+c^2)+5(ab+bc+ac)}\geq (\frac{1}{2}+\frac{1}{6})(a^2+b^2+c^2)=RHS</math>, our proof is done | ||
| + | |||
| + | ~bluesoul | ||
==See Also== | ==See Also== | ||
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{{USAJMO newbox|year=2012|num-b=2|num-a=4}} | {{USAJMO newbox|year=2012|num-b=2|num-a=4}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 19:20, 17 June 2022
Contents
Problem
Let 
, 
, 
be positive real numbers. Prove that
![\[\frac{a^3 + 3b^3}{5a + b} + \frac{b^3 + 3c^3}{5b + c} + \frac{c^3 + 3a^3}{5c + a} \ge \frac{2}{3} (a^2 + b^2 + c^2).\]](http://latex.artofproblemsolving.com/4/4/8/448b311f5688c5c18e520bfe1a9c56906700c3d2.png)
Solution
By the Cauchy-Schwarz inequality,
![\[[a(5a + b) + b(5b + c) + c(5c + a)] \left( \frac{a^3}{5a + b} + \frac{b^3}{5b + c} + \frac{c^3}{5c + a} \right) \ge (a^2 + b^2 + c^2)^2,\]](http://latex.artofproblemsolving.com/c/8/5/c853ed04ce97a5d53eb2c179cc962c2d9e0af755.png)
so
![\[\frac{a^3}{5a + b} + \frac{b^3}{5b + c} + \frac{c^3}{5c + a} \ge \frac{(a^2 + b^2 + c^2)^2}{5a^2 + 5b^2 + 5c^2 + ab + ac + bc}.\]](http://latex.artofproblemsolving.com/c/7/0/c70b85f9d08ce18cabc92505778450fa22084b5d.png)
Since 
,
![\[\frac{(a^2 + b^2 + c^2)^2}{5a^2 + 5b^2 + 5c^2 + ab + ac + bc} \ge \frac{(a^2 + b^2 + c^2)^2}{6a^2 + 6b^2 + 6c^2} = \frac{1}{6} (a^2 + b^2 + c^2).\]](http://latex.artofproblemsolving.com/c/6/7/c67539d05946214142c90df1baebf0bcc6fc3875.png)
Hence,
![\[\frac{a^3}{5a + b} + \frac{b^3}{5b + c} + \frac{c^3}{5c + a} \ge \frac{1}{6} (a^2 + b^2 + c^2).\]](http://latex.artofproblemsolving.com/2/5/b/25b8b46bbae4a04cad9f782369cbd6b469179254.png)
Again by the Cauchy-Schwarz inequality,
![\[[b(5a + b) + c(5b + c) + a(5c + a)] \left( \frac{b^3}{5a + b} + \frac{c^3}{5b + c} + \frac{a^3}{5c + a} \right) \ge (a^2 + b^2 + c^2)^2,\]](http://latex.artofproblemsolving.com/d/f/4/df4ba3115174961740e725bbf3e055ee3c95ca60.png)
so
![\[\frac{b^3}{5a + b} + \frac{c^3}{5b + c} + \frac{a^3}{5c + a} \ge \frac{(a^2 + b^2 + c^2)^2}{a^2 + b^2 + c^2 + 5ab + 5ac + 5bc}.\]](http://latex.artofproblemsolving.com/a/0/9/a091852266f2a5080dcdc2d9bd4f5aabb253c303.png)
Since ,
![\[\frac{(a^2 + b^2 + c^2)^2}{a^2 + b^2 + c^2 + 5ab + 5ac + 5bc} \ge \frac{(a^2 + b^2 + c^2)^2}{6a^2 + 6b^2 + 6c^2} = \frac{1}{6} (a^2 + b^2 + c^2).\]](http://latex.artofproblemsolving.com/0/c/5/0c593c978f6e27800e4421768d0df18ee5d7ee1d.png)
Hence,
![\[\frac{b^3}{5a + b} + \frac{c^3}{5b + c} + \frac{a^3}{5c + a} \ge \frac{1}{6} (a^2 + b^2 + c^2).\]](http://latex.artofproblemsolving.com/5/4/a/54abb3c7c80aa1733717f8d78fa9680a3cf52dc2.png)
Therefore,
![\[\frac{a^3 + 3b^3}{5a + b} + \frac{b^3 + 3c^3}{5b + c} + \frac{c^3 + 3a^3}{5c + a} \ge \frac{1 + 3}{6} (a^2 + b^2 + c^2) = \frac{2}{3} (a^2 + b^2 + c^2).\]](http://latex.artofproblemsolving.com/9/7/3/9736e742df8602ada135197c276e1cdd5c4d108c.png)
Solution 2
Titu's Lemma: The sum of multiple fractions in the form 
where 
and 
are sequences of real numbers is greater than of equal to the square of the sum of all 
divided by the sum of all 
, where i is a whole number less than n+1. Titu's Lemma can be proved using the Cauchy-Schwarz Inequality after multiplying out the denominator of the RHS.
Consider the LHS of the proposed inequality. Split up the numerators and multiply both sides of the fraction by either a or 3a to make the LHS
![\[\sum_{cyc} \frac {a^4} {5a^2+ab}+\sum_{cyc} \frac {9a^4} {15ac+3a^2}\]](http://latex.artofproblemsolving.com/b/b/5/bb59c991eeec060f418f26510fdcedc497241af0.png)
(Cyclic notation is a special notation in which all permutations of (a,b,c) is the summation command.)
Then use Titu's Lemma on all terms: ![\[\sum_{cyc} \frac {a^4} {5a^2+ab} +\sum_{cyc} \frac {9a^4} {15ac+3a^2} \ge \frac {16(a^2+b^2+c^2)^2} {8(a^2+b^2+c^2)+16(ab+bc+ca)} \ge \frac {16(a^2 + b^2 + c^2)^2}{24(a^2 + b^2 + c^2)} = \frac{2}{3} (a^2 + b^2 + c^2)\]](http://latex.artofproblemsolving.com/1/e/6/1e6ffd148cf766ae878225a88cdfbd6a8a32a8ac.png)
owing to the fact that 
, which is actually equivalent to 
!
Solution 3
We proceed to prove that
![\[\frac{a^3 + 3b^3}{5a + b} \ge -\frac{1}{36} a^2 + \frac{25}{36} b^2\]](http://latex.artofproblemsolving.com/c/5/2/c52d3a698663e0f197adc851cb5e2dae41a640ab.png)
(then the inequality in question is just the cyclic sum of both sides, since ![\[\sum_{cyc} (-\frac{1}{36} a^2 + \frac{25}{36} b^2) = \frac{24}{36}\sum_{cyc} a^2 = \frac{2}{3} (a^2+b^2+c^2)\]](http://latex.artofproblemsolving.com/8/5/b/85b9c8f86b494c420b447b4ab18833e3c7888935.png)
)
Indeed, by AP-GP, we have
![\[41 (a^3 + b^3+b^3) \ge 41 \cdot 3 ab^2\]](http://latex.artofproblemsolving.com/0/4/0/040230a38420df1ebd9c6f4a3b8ed4864046a31c.png)
and
![\[b^3 + a^2b \ge 2 ab^2\]](http://latex.artofproblemsolving.com/7/8/c/78cc2657bcc406488570de8825696deaaf82a955.png)
Summing up, we have
![\[41a^3 + 83b^3 + a^2 b \ge 125 ab^2\]](http://latex.artofproblemsolving.com/c/c/6/cc6c8baaba46e8a756ed5f46223336f7a3205ad8.png)
which is equivalent to:
![\[36(a^3 + 3b^3) \ge (5a + b)(-a^2 + 25b^2)\]](http://latex.artofproblemsolving.com/5/2/4/524e82b05d73c8c604d6457527696244b5eda9d0.png)
Dividing 
from both sides, the desired inequality is proved.
--Lightest 15:31, 7 May 2012 (EDT)
Solution 4
By Cauchy-Schwarz,
![\[\left(\sum_{cyc} \dfrac{a^3}{5a + b} + \dfrac{b^3}{5a + b} + \dfrac{b^3}{5a + b} + \dfrac{b^3}{5a + b} \right) \ge \dfrac{\left( \sum_{cyc} a^2 + b^2 + b^2 + b^2 \right)^2}{ \left( \sum_{cyc} a(5a + b) + b(5a + b) + b(5a + b) + b(5a + b) \right) }\]](http://latex.artofproblemsolving.com/5/d/d/5dd205430dfa38b6aa7aab8207b972ba9b392cfc.png)
![\[= \dfrac{\left(4a^2 + 4b^2 + 4c^2\right)^2}{\left(8a^2 + 8b^2 + 8c^2\right) + \left(16ab + 16bc + 16ca\right)}\]](http://latex.artofproblemsolving.com/c/d/1/cd12be6cc3fab31909d40c6674da42903db894d6.png)
![\[\ge \dfrac{16\left(a^2 + b^2 + c^2\right)^2}{\left(8a^2 + 8b^2 + 8c^2\right) + \left(8a^2 + 8b^2\right) + \left(8b^2 + 8c^2\right) + \left(8c^2 + 8a^2\right)}\]](http://latex.artofproblemsolving.com/6/c/1/6c1a2614c369fe54215fbcd0bcdce724ff9e12c3.png)
(by AM-GM) ![\[= \dfrac{2}{3}\left(a^2 + b^2 + c^2\right)\]](http://latex.artofproblemsolving.com/b/9/c/b9c7c7316e0057962340f7896f96616b0757235b.png)
as desired.
Solution 5
We note that if we can prove that there exists a real number 
such that ![\[\frac{a^3+3b^3}{5a+b}\geq xa^2+(\frac{2}{3}-x)b^2\]](http://latex.artofproblemsolving.com/9/f/c/9fc8cc7b68f1a1d227dd9c2f94bcd46b8f28bd24.png)
for all positive reals and , then we are done since both sides are a cyclic sum of 
, and . Note that if we multiply both and by a constant 
, the left and right sides will both multiply by 
(since the numerator on the left will multiply by 
and the denominator will multiply by . This means that if the inequality is satisfied for an ordered pair 
, then it is also satisfied for 
for any positive real . Thus, without loss of generality we can let 
,and our inequality becomes ![\[\frac{a^3+3}{5a+1}\geq a^2x+\frac{2}{3}-x.\]](http://latex.artofproblemsolving.com/0/6/e/06e6718a41c304a5f8a7c598007fbd66bf40b5fc.png)
Expanding this and rearranging we get 
. Since 
is a root of the left side, we can factor this as ![\[(a-1)(a^2(1-5x)+a(1-6x)-x-\frac{7}{3})\geq0.\]](http://latex.artofproblemsolving.com/5/2/0/520f541c8897c940cba7e9c8c253fc7bda812e79.png)
Note that for this to always be nonnegative, we must have 
be positive when 
and negative when 
. Thus, it must have a root at 
, so we plug in is a root and find that we must have 
. When , the expression factors as 
which obviously is nonnegative for all positive reals , so we are done.
~john0512, bronzetruck2016
Solution 6
which tells that 
, our proof is done
~bluesoul
See Also
| 2012 USAJMO (Problems • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 • 6 | ||
| All USAJMO Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
